Chat（推論モード）

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                "content": "The sum of all integers from 1 to 100 is:\n\n\\[\n1 + 2 + 3 + \\cdots + 100 = 5050\n\\]\n\nThis result is calculated using the formula for the sum of the first \\( n \\) natural numbers:\n\n\\[\nS = \\frac{n(n+1)}{2}\n\\]\n\nwhere \\( n = 100 \\). Substituting, we get:\n\n\\[\nS = \\frac{100 \\times 101}{2} = 5050\n\\]",
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                "content": "The sum of all integers from 1 to 100 is:\n\n\\[\n1 + 2 + 3 + \\cdots + 100 = 5050\n\\]\n\nThis result is calculated using the formula for the sum of the first \\( n \\) natural numbers:\n\n\\[\nS = \\frac{n(n+1)}{2}\n\\]\n\nwhere \\( n = 100 \\). Substituting, we get:\n\n\\[\nS = \\frac{100 \\times 101}{2} = 5050\n\\]",
                "reasoning_content": "First, the problem is to find the sum of all integers from 1 to 100. That's 1 + 2 + 3 + ... + 100.\n\nI recall that there's a formula for the sum of the first n natural numbers. The formula is:\n\n\\[ S = \\frac{n(n+1)}{2} \\]\n\nHere, n is 100. So, I plug in n=100 into the formula.\n\n\\[ S = \\frac{100 \\times (100 + 1)}{2} = \\frac{100 \\times 101}{2} \\]\n\nNow, I calculate that. 100 times 101 is 10,100. Then, divide by 2: 10,100 / 2 = 5,050.\n\nSo, the sum should be 5,050.\n\nI remember that the story about Gauss solving this quickly as a kid. He paired up numbers: 1+100=101, 2+99=101, 3+98=101, and so on. Since there are 100 numbers, that's 50 pairs (because 100/2=50). Each pair sums to 101, so 50 times 101 is 5,050. Same answer.\n\nSo, I'm confident with the result.The sum of all integers from 1 to 100 is calculated using the formula for the sum of an arithmetic series:\n\n\\[ S = \\frac{n(n+1)}{2} \\]\n\nwhere \\( n = 100 \\).\n\nSubstituting the value:\n\n\\[ S = \\frac{100 \\times 101}{2} = \\frac{10100}{2} = 5050 \\]\n\nThus, \\( 1 + 2 + 3 + \\cdots + 100 = 5050 \\).",
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